Question

1/(1 - sinx) - 1/(1 + sinx)

I believe the answer is 1/cos^2x, but I don't know why. Can someone help walk me through the steps?

Answer 1

\[\large\rm \frac{1}{1-\sin x}-\frac{1}{1+\sin x}\]

Do you remember your Pythagorean Identity involving sine and cosine? :)
Here is one way we can write that relationship,\[\large\rm 1-\sin^2x=\cos^2x\]

Answer 2

Notice that the left side of this identity is really the `difference of two squared numbers`\[\large\rm 1^2-\sin^2x\]This will factor into conjugates,\[\large\rm =(1-\sin x)(1+\sin x)\]

Answer 3

That should give you a sense of what direction we'll be heading in our steps :)

Answer 4

If we're able to get some (1-sin x)(1+sin x) in our denominators,
we'll be able to apply our Pythagorean Identity.

Answer 5

So if that's the case,
I guess we would like to have a (1+sin x) in the bottom and top of our first fraction,\[\large\rm \color{royalblue}{\frac{1+\sin x}{1+\sin x}}\cdot\frac{1}{1-\sin x}-\frac{1}{1+\sin x}\]

Answer 6

Which becomes,\[\large\rm \frac{1+\sin x}{(1+\sin x)(1-\sin x)}-\frac{1}{1+\sin x}\]

Answer 7

What do you think gates?
Too confusing?

Answer 8

It's fairly confusing. That's probably why I didn't quite absorb the information in class. This whole chapter is crazy.

Answer 9

\[\frac{1}{1-b}-\frac{1}{1+b}\]
\[=\frac{1+b-(1-b)}{1-b^2}=\frac{2}{1-b^2}\] it is 90% algebra

Answer 10

Well you're familiar with the idea of a `common denominator`, yes?\[\large\rm \frac{1}{a}-\frac{1}{b}\quad=\quad \frac{b}{ab}-\frac{a}{ab}\quad=\frac{b-a}{ab}\]That's a good starting point for this problem.

We would like a new denominator which has both denominators as factors.
So our new denominator should consist of both (1-sin x) and (1+sin x).
So we go from here,\[\large\rm \frac{1}{(1-\sin x)}-\frac{1}{(1+\sin x)}\]to here,\[\large\rm \frac{\color{royalblue}{(1+\sin x)}}{\color{royalblue}{(1+\sin x)}(1-\sin x)}-\frac{\color{orangered}{(1-\sin x)}}{(1+\sin x)\color{orangered}{(1-\sin x)}}\]

Answer 11

Yeah so this all leads back down to 1/1-sinx - 1/1+sin, but where do I go from there?

Answer 12

Well we don't want to go backwards.
We're turning it into this big mess for a specific reason.
Now we're able to combine it into a single fraction,\[\large\rm \frac{(1+\sin x)-(1-\sin x)}{(1+\sin x)(1-\sin x)}\]

Answer 13

If we drop the brackets in the numerator, distributing the negative,\[\large\rm \frac{1+\sin x-1+\sin x}{(1+\sin x)(1-\sin x)}\]We can combine like-terms.
Looks like the 1's will cancel out, ya?

Answer 14

Yup

Answer 15

\[\large\rm \frac{2\sin x}{(1+\sin x)(1-\sin x)}\]So that simplifies down the top as far as we're able to go.

Answer 16

And again, bottom is an identity.
If you don't quite understand it,
maybe just "believe it" for right now.\[\large\rm (1+\sin x)(1-\sin x)=\cos^2x\]

Answer 17

\[\large\rm \frac{2\sin x}{\cos^2x}\]

Answer 18

Oh okay, thanks so much!

Answer 19

np